Yes, it is independent of that choice because $\delta_\epsilon$ always converges weakly to the pullback to the small diagonal $\Delta_n$.

Honestly, I find it amusing (and concerning) that Bergh-Löfström are so cavalier with the use of the Bochner integral, for e.g. math.stackexchange.com/questions/2880662/…

(continued) I have the impression that the discrete method does carry through to $E_0,E_1$ just normed, but that of course has to be checked. However, when it comes to considering admissible operators $T$, the commutation of $T$ with the Bochner integral is tacitly used by Adams-Fournier as well (see Theorem 7.23, pp. 220-221 of ibid.). One would need first to try combining both theorems just quoted to see how the proof of the latter theorem carries through in the discrete version of the $J$-method, and then check if the ensuing argument carries through to the normed case.

The exposition of the $J$-method in the book by R. A. Adams and J. J. F. Fournier, Sobolev Spaces (Elsevier, 2002) indeed assumes $E_0$ and $E_1$ to be Banach spaces and yes, the Bochner integral is in fact used therein. There is a discrete version of the $J$-method which circumvents the use of the Bochner integral (see e.g. Theorem 7.15, pp. 213-215 of ibid.), but I'm not sure if it can be extended to $E_0,E_1$ just normed. Btw, the Adams-Fournier exposition is mainly based on the book by P. L. Butzer and H. Berens, Semi-Groups of Operators and Approximation (Springer, 1967).

Yes, for $:\widehat{\mathcal{O}}:=:\!\phi^{n_1}\!\!:\!(x)$ in this case. Everything between colons is a single field operator-valued distribution on $\mathbb{R}^4$, thus depending on a single space-time argument.

Vev's of Wick powers are zero, that's just the case $k=1$ in my last comment above, $n_1$ can be any natural number. I still believe you're mistaking the Wick power itself for the number of Wick monomials in the vev.

Sorry, I'll expand my explanation. Formula (16) requires that all variables $x_1,\ldots,x_k$ in the vev $\langle\Omega,:\!\phi^{n_1}\!\!:\!(x_1)\cdots:\!\phi^{n_k}\!\!:\!(x_k)\Omega\rangle$ should be paired among themselves in the sum in the rhs. Particularly, this implies that $k$ should be even in order for one to have any term in that sum (just recall the definition of pairing in my answer above). If $k$ is odd (particularly if $k=1$), there are no terms in the sum in the rhs and thus that sum is zero. Maybe you are mistaking the powers $n_1,\ldots,n_k$ for the number $k$ of such powers.

More generally, the vacuum expectation value of a product of any odd number of Wick powers of $\phi$ is zero for the same reason, generalizing the analogous result for $\phi$ itself.

There is no contradiction, the vacuum expectation value of any Wick power of $\phi$ does yield zero as it should. In the language of formula (16), this means there is no possible pairing of variables simply because there is just one variable, so the sum is trivially zero. When I discussed smoothness of the matrix elements of $:\!\phi^n\!\!:\!(x)$ in my answer, I did mention that the given form of those was as described there whenever this matrix element is not zero. Vacuum expectation values is precisely one of the cases where the result is zero.

(Continued) Provided, of course, you take the limits involved in the i0 prescription in the correct order. Agaim, this is similar to what happens here, recalling that the product should be seen as the pullback of the tensor product by the diagonal map in order to be extended to distributions satisfying Hörmander's wave front set criterion.

@MartinHairer the precise shape of the singularities matters here, that's where Hörmander's wave front set criterion for the existence of a product enters. As a simple example, $\frac{1}{x^n}$ doesn't define a distribution on $\mathbb{R}$ for it's not locally integrable at 0, but $\frac{1}{(x+i0)^n}$ does for all $n\in\mathbb{n}$ for the same reason as $\omega_2(x,y)^n$.

@Isaac yes, the situation here may be seen as an unwarranted interchange of limits, that's why the Wick product has to be defined in the way I described and not as you've suggested.